YES

The TRS could be proven terminating. The proof took 24 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (11ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

f#(s(x), s(y), z, u)f#(s(x), minus(y, x), z, u)f#(s(x), 0, z, u)f#(x, u, minus(z, s(x)), u)
f#(s(x), s(y), z, u)f#(x, u, z, u)perfectp#(s(x))f#(x, s(0), s(x), s(x))

Rewrite Rules

perfectp(0)falseperfectp(s(x))f(x, s(0), s(x), s(x))
f(0, y, 0, u)truef(0, y, s(z), u)false
f(s(x), 0, z, u)f(x, u, minus(z, s(x)), u)f(s(x), s(y), z, u)if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Original Signature

Termination of terms over the following signature is verified: f, 0, minus, s, le, if, perfectp, false, true

Strategy


The following SCCs where found

f#(s(x), 0, z, u) → f#(x, u, minus(z, s(x)), u)f#(s(x), s(y), z, u) → f#(x, u, z, u)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

f#(s(x), 0, z, u)f#(x, u, minus(z, s(x)), u)f#(s(x), s(y), z, u)f#(x, u, z, u)

Rewrite Rules

perfectp(0)falseperfectp(s(x))f(x, s(0), s(x), s(x))
f(0, y, 0, u)truef(0, y, s(z), u)false
f(s(x), 0, z, u)f(x, u, minus(z, s(x)), u)f(s(x), s(y), z, u)if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Original Signature

Termination of terms over the following signature is verified: f, 0, minus, s, le, if, perfectp, false, true

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

f#(s(x), s(y), z, u)f#(x, u, z, u)f#(s(x), 0, z, u)f#(x, u, minus(z, s(x)), u)